王予柔月事妹是哪一集:在等差数列中,若Sm/Sn=m^2/n^2(m不等于n),则am/an=

来源:百度文库 编辑:高校问答 时间:2024/03/29 15:22:53
答案是(2m-1)/(2n-1)
要详细的过程,谢谢!

原问题即 : 有两个数列,{An} {Bn}, 若 Sm : Sn = m^2 : n^2
求 Am : Bn (公差分别为d1, d2)
解: Sm = A1d + 0.5*m(m-1)d1 = 0.5d1m^2 + m(A1-0.5d1)
Sn = 0.5d2n^2 + n(B1-0.5d2)
∵ Sm : Sn = m^2 : n^2 ∴ A1-0.5d1=0 B1-0.5d2=0
Sm = 0.5d1m^2 Sn = 0.5d2n^2 ∴d1=d2=d A1=A2=0.5d
∵ Am = A1 + (m-1)d = d(m-0.5) Bn = d(n-0.5)
∴ Am : Bn = (m-0.5) : (n-0.5) = (2m-1) : (2n-1)

Let T_m=S_m/m^2.

Then T_m=T_n , for all m,n

So T_m=T_1=S_1=a_1, that is, S_m=a_1*m^2.

Hence a_m=a_1(m^2-(m-1)^2)=a_1(2m-1).

Thus we have a_m/a_n=(2m-1)/(2n-1)